Answer Radium & Town Population: 200yrs, 800k

[bergausstein]

1. Radium decomposes at a rate proportional to the amount at any instant. In 100 years, 100mg of radium decomposes to 96mg. How many mg will be left after 200 years?

2. if a population of a town doubled in the past 25 years and the present population is 300,000 when will the town have a population of 800,000?prob 1.

since 3mg of 100mg radium have decomposed over a period of 100 years this amount is 3% of the original amount.$\frac{R_0-0.03R_0}{R_0}=\frac{R_0\,e^{k100}}{R_0}$

$\ln(1-0.03)=\ln(e^{k100})$

$\ln(1-0.03)=100k$

$k=\frac{\ln(1-0.03)}{100}$

when t=200

$R(200)=R_0\,e^{\ln(1-0.03)^2}$

$R(200)=100(0.9409)$

$R=94.09mg$ is this correct?

prob 2

$\frac{dP}{dt}=kP$

$P(t)=P_0\,e^{kt}$

when t=25; $P_0=2P_0$

$\frac{2P_0}{P_0}=\frac{P_0\,e^{25k}}{P_0}$

$2=e^{25k}$

$\ln2=25k$

$k=\frac{\ln2}{25}$$\frac{dP}{dt}=kR$

$P(t)=P_0\,e^{kt}$

$P(t)=P_0\,e^{\frac{\ln2}{25}t}$

$800,000=300,000(2^{\frac{t}{25}}$

$\ln2.67=\frac{t}{25}\ln2$

$t= 35.42$years

is this correct?

 

[HallsofIvy]

1. Radium decomposes at a rate proportional to the amount at any instant. In 100 years, 100mg of radium decomposes to 96mg. How many mg will be left after 200 years?

2. if a population of a town doubled in the past 25 years and the present population is 300,000 when will the town have a population of 800,000?prob 1.

since 3mg of 100mg radium have decomposed over a period of 100 years this amount is 3% of the original amount.$\frac{R_0-0.03R_0}{R_0}=\frac{R_0\,e^{k100}}{R_0}$

Where did ".03" come from? The problem said that amount reduced from 100 to 96 mg. Where did the "3 mg" you refer to come from?
100-96= 4.

$\ln(1-0.03)=\ln(e^{k100})$

$\ln(1-0.03)=100k$

$k=\frac{\ln(1-0.03)}{100}$

when t=200

$R(200)=R_0\,e^{\ln(1-0.03)^2}$

$R(200)=100(0.9409)$

$R=94.09mg$ is this correct?

prob 2

$\frac{dP}{dt}=kP$

$P(t)=P_0\,e^{kt}$

when t=25; $P_0=2P_0$

$\frac{2P_0}{P_0}=\frac{P_0\,e^{25k}}{P_0}$

$2=e^{25k}$

$\ln2=25k$

$k=\frac{\ln2}{25}$$\frac{dP}{dt}=kR$

$P(t)=P_0\,e^{kt}$

$P(t)=P_0\,e^{\frac{\ln2}{25}t}$

$800,000=300,000(2^{\frac{t}{25}}$

$\ln2.67=\frac{t}{25}\ln2$

$t= 35.42$years

is this correct?

When I do it, using 8/3= 2.66666666666666667, I get 35.38 years. Don't round off until you have to! But the data given: 800,000, 300,000, and 25 are at most to 2 significant figures so I would say answer years anyway.

 

[MarkFL]

If we're going to actually solve an initial value problem, we could develop the general formula as follows:

\(\displaystyle \frac{dy}{dt}=ky\) where \(\displaystyle y(0)=y_0\)

Separate variables, integrate with respect to $t$, switch dummy variables for clarity, and use the given boundaries as limits:

\(\displaystyle \int_{y_0}^{y(t)}\frac{1}{u}\,du=k\int_0^t\,dv\)

Apply the FTOC:

\(\displaystyle \ln\left|\frac{y(t)}{y_0} \right|=kt\)

Now, if we are given another point on the solution $\left(a,y_a \right)$ then we have:

\(\displaystyle \ln\left|\frac{y_a}{y_0} \right|=ka\implies k=\frac{1}{a}\ln\left|\frac{y_a}{y_0} \right|\)

And so we find:

\(\displaystyle \ln\left|\frac{y(t)}{y_0} \right|=\frac{1}{a}\ln\left|\frac{y_a}{y_0} \right|t\)

Hence:

(1) \(\displaystyle y(t)=y_0\left(\frac{y_a}{y_0} \right)^{\frac{t}{a}}\)

(2) \(\displaystyle t=\frac{a\ln\left|\frac{y(t)}{y_0} \right|}{\ln\left|\frac{y_a}{y_0} \right|}\)

Now we have formulas to answer this type of problem.

Problem 1: We want to use (1).

We identify:

\(\displaystyle y_0=100\text{ mg},\,y_a=96\text{ mg},\,a=100\text{ yr},\,t=200\text{ yr}\)

Plugging this into (1), we obtain:

\(\displaystyle y(t)=100\left(\frac{96}{100} \right)^{\frac{200}{100}}\text{ mg}=100\left(\frac{24}{25} \right)^2\text{ mg}=\frac{2304}{25}\text{ mg}=92.16\text{ mg}\)

Your error was, as HallsofIvy pointed out, one of subtraction.

Problem 2: We want to use (2).

We identify:

\(\displaystyle a=25\text{ yr},\,y_0=150000\,y_a=300000,\,y(t)=800000\)

Plugging this into (2), we obtain:

\(\displaystyle t=\frac{25\ln\left|\frac{800000}{150000} \right|}{\ln\left|\frac{300000}{150000} \right|}\text{ yr}=\frac{25\ln\left(\frac{16}{3} \right)}{\ln\left(2 \right)}\text{ yr}\)

Now, we want to subtract 25 from this because the starting point is 25 years ago. So we find the population will be 800,000 approximately 35.3759374819711 years from now.